I've been learning Set Theory from both Danel Cunningham and Herbert Enderton's books and I can't wrap my head around how they prove the Recursion theorem on $\omega$:
Let $A$ be a set and $a \in$ A. Suppose that $f : A → A$ is a function. Then there exists a unique function $h : ω → A$ such that
(1) $h(0) = a$,
(2) $h(n + ) = f (h(n))$, for all n $\in$ ω.
Both of them starts off the proof by constructing a set that is contains all the acceptable functions i.e. satisfy the above $2$ conditions.
My question is why is it even necessary to have a set of functions in the first place? Why don't just assume a relationship $R \subset \omega \times A $ that satisfies the $2$ condition (using Subset Axiom) and work from there?
I do understand the basic ZF axioms (Extensionality, Empty Set, Pairing, Union, Power Set, Subset) that I deem handy when solving some other problems but not this set of functions thingy. Is this supposed to be a proving trick or some logical connection that I missed out on? Or am I not understanding the basic ZF axioms fully?
From that point onwards, I could understand the proofs fully, but it leaves me with much insecurity and anxiety that I don't have good enough intuition or the smarts to flow along with a textbook because I do believe there must be some kind of intuition that this approach is built on top of.
Note that the difficulty of the proof is to find the existence of the relation. You can only assume that $R$ exists (and is a function with the $2$ properties) once you are able to show this...
Maybe you want to use something else to show the existence: What do you mean be the "Subset axiom"?
Maybe you mean the subset schema of specification/comprehension?
But for that you need to describe a property $F$ such that $(n,a)\in R\iff F(n,a)$. Note that the description of the property $F$ cannot refer to $R$, because you want to use this property in the comprehension schema to define $R$ in the first place...